77.2. SOME FUNDAMENTAL THEOREMS 2605

Lemma 77.2.12 The mapping ω → Γn (ω) is an F measurable set-valued map with val-ues in X. If σ is a measurable selection, then for each t, ω → σ (t,ω) is F measurableand (t,ω)→ σ (t,ω) is B ([0,T ])×F measurable.

We note that if σ is a measurable selection then σ (ω) ∈ Γn (ω), so σ = σ (·,ω) is acontinuous function. To have σ measurable would mean that σ

−1k (open) ∈F , where the

open set is in C ([0,T ]).Proof: Let O be a basic open set in X . Then O = ∏

∞k=1 Ok, where Ok is a proper open

set of C ([0,T ]) only for k ∈ {k1, · · · ,kr}. Thus there is a proper open set in these positionsand in every other position the open set is the whole space C ([0,T ]) .We need to show that

Γn− (O)≡ {ω : Γ

n (ω)∩O ̸= /0} ∈F .

Now, Γn− (O) = ∩ri=1

{ω : Γn (ω)ki

∩Oki ̸= /0}

, so we consider whether{ω : Γ

n (ω)ki∩Oki ̸= /0

}∈F . (77.2.4)

From the definition of Γn (ω) , this is equivalent to the condition that

fki (XNC (ω)u j (·,ω)) = (f(XNC (ω)u j (·,ω)))ki∈ Oki

for some j ≥ n, and so the set in 77.2.4 is of the form

∪∞j=n

{ω : (f(XNC (ω)u j (·,ω)))ki

∈ Oki

}.

Now ω → (f(XNC (ω)u j (·,ω)))kiis F measurable into C ([0,T ]) and so the above set is

in F . To see this, let g ∈C ([0,T ]) and consider the inverse image of the ball with radius rand center g,

B(g,r) ={

ω :∥∥∥(XNC (ω) f(u j (·,ω)))ki

−g∥∥∥

C([0,T ])< r}.

By continuity considerations,∥∥∥(XNC (ω) f(u j (·,ω)))ki−g∥∥∥

C([0,T ])

= supt∈Q∩[0,T ]

∣∣∣(XNC (ω) f(u j (t,ω)))ki−g(t)

∣∣∣ ,which is the sup over countably many F measurable functions. Thus, it is F measurable.Since every open set is the countable union of such balls, it follows that the claim about Fmeasurability is valid. Hence, Γn− (O) is F measurable whenever O is a basic open set.

Now, X is a separable metric space and so every open set is a countable union of thesebasic sets. Let U ⊆ X be open with U = ∪∞

l=1Ol where Ol is a basic open set as above.Then,

Γn− (U) = ∪∞

l=1Γn− (Ol) ∈F .